## Calculus: Early Transcendentals 8th Edition

$$y=-x+\pi$$
Finding the derivative of the curve: $y'=\frac{d\sin(\sin{x})}{dx}$ Using the chain rule: $y'=\frac{d\sin(\sin{x})}{d\sin{x}} \times \frac{d\sin{x}}{dx}$ $=\cos(\sin{x}) \times\cos{x}$ Finding the slope ($m$) to the tangent line at ($\pi$,0): $m=y'(\pi)$ $=\cos(\sin{\pi}) \times\cos{\pi}=\cos(0)\times(-1)=1\times(-1)=-1$ From the point-slope form of a linear equation: $y-0=-1(x-\pi)$ Thus, an equation of the tangent line would be: $y=-x+\pi$